Question: $h(x)=\begin{cases} 5x&\text{for }x<-2 \\\\ x^3-2&\text{for }x\geq-2 \end{cases}$ Find $\lim_{x\to -2}h(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-10$ (Choice B) B $-2$ (Choice C) C $10$ (Choice D) D The limit doesn't exist.
$x=-2$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to -2}h(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $-2$ from the left. We will use the fact that $h(x)=5x$ for $x$ -values smaller than $-2$. $\begin{aligned} &\phantom{=}\lim_{x\to -2^-}h(x) \\\\ &=\lim_{x\to -2^-}5x \\\\ &=5(-2)&\gray{\text{Direct substitution}} \\\\ &=-10 \end{aligned}$ Let's find the limit as $x$ approaches $-2$ from the right. We will use the fact that $h(x)=x^3-2$ for $x$ -values greater than $-2$. $\begin{aligned} &\phantom{=}\lim_{x\to -2^+}h(x) \\\\ &=\lim_{x\to -2^+}x^3-2 \\\\ &=(-2)^3-2&\gray{\text{Direct substitution}} \\\\ &=-10 \end{aligned}$ The one-sided limits are both equal to $-10$. This means that $\lim_{x\to -2}h(x)=-10$.